Why is free field called free ?

[kof9595995]

Why can't we just use the QM operator directly,but have to map it to a new object? You mean the 1-particle state in QFT is different from the 1- particle state in QM? How?

 

[A. Neumaier]

Why can't we just use the QM operator directly,but have to map it to a new object? You mean the 1-particle state in QFT is different from the 1- particle state in QM? How?

The 1-particle state spaces are isomorphic.

But second quantization allows the 1-particle operators to act on arbitrary N-particle states, and on states of indefinite particle number.

 

[kof9595995]

About second quantization, I guess what really confused me is why the field operator equation in QFT look so much like the equation for wavefunction in QM, despite that we interpret them very differently.
To make the question clearer:
A generic one-particle state can be written as
|\Phi > = \int {\Phi (x,t)|x > dx}
So Phi(x) is the wavefunction
With field operators we can write it as
|\Phi > = \int {\Phi (x,t){\psi _{op}}(x)dx} |0 >
Psi_op is the field operator which creates a particle at x.
Now QM requires some equation to be satified (Schrodinger for example):
- \frac{\hbar }{{2m}}\frac{{{\partial ^2}}}{{\partial {x^2}}}\Phi (x,t) = i\hbar \frac{\partial }{{\partial t}}\Phi (x,t)...(1)
But in the corresponding field theory we also have
- \frac{\hbar }{{2m}}\frac{{{\partial ^2}}}{{\partial {x^2}}}{\psi _{op}}(x,t) = i\hbar \frac{\partial }{{\partial t}}{\psi _{op}}(x,t)...(2)
It confused me that they look so much alike, I know the way we construct psi_op naturally makes it satisfy (2), but this only makes me feel like it's a pure coincidence. I would be happy if we can somehow show (1) and (2) is indeed logically connected.

 

[kof9595995]

Operators A(p,q) on the 1-particle Hilbert space extend to corresponding field operators by means of the functor that maps A to a bilinear form N(A) in the c/a operators, which in the simplest case of a scalar field takes the form
N(A)=\int dp a^*(p)A(p,q)a(p).
[for some unknown reason, the dp prints as dx on my screen. Note:q is the differential operator in the momentum representation, not a variable.]

I used this prescription and try to calculate commutator [x,p], but it seems to lead me nowhere:
\begin{array}{l}<br /> [x,p]=\int {dpdk\{ {a^\dag }(p)\frac{\partial }{{\partial p}}} a(p)k{a^\dag }(k)a(k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} ......(1) \\ <br /> = i\int {dpdk\{ k{a^\dag }(p)\frac{\partial }{{\partial p}}} [{a^\dag }(k)a(p) + \delta (p - k)]a(k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} ....(2) \\ <br /> = i\int {dpdk\{ k{a^\dag }(k){a^\dag }(p)\frac{\partial }{{\partial p}}} a(p)a(k) + k{a^\dag }(p)\frac{\partial }{{\partial p}}\delta (p - k)a(k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} ...(3) \\ <br /> = i\int {dpdk\{ k{a^\dag }(k){a^\dag }(p)a(k)\frac{\partial }{{\partial p}}} a(p) + k{a^\dag }(p)a(k)\frac{\partial }{{\partial p}}\delta (p - k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} ...(4) \\ <br /> = i\int {dpdk\{ k{a^\dag }(k)[a(k){a^\dag }(p) - \delta (k - p)]\frac{\partial }{{\partial p}}} a(p) + k{a^\dag }(p)a(k)\frac{\partial }{{\partial p}}\delta (p - k) - k{a^\dag }(k)a(k){a^\dag }(p)\frac{\partial }{{\partial p}}a(p)\} ...(5) \\ <br /> = i\int {dpdk\{ - k{a^\dag }(k)\delta (k - p)\frac{\partial }{{\partial p}}} a(p) + k{a^\dag }(p)a(k)\frac{\partial }{{\partial p}}\delta (p - k)\} ......(6) \\ <br /> \end{array}
Did I do something wrong?(Sorry, I know it looks horribly tedious)

 

[A. Neumaier]

I used this prescription and try to calculate commutator [x,p], (Sorry, I know it looks horribly tedious)

In the definiition of q you forgot a factor of =i/hbar (or -i/hbar).

Ding these calculations by brute force is tedious indeed. But it can be done in a smarter way:

First compute for general A the commutator [N(A),a(p)]. Then conjugate the resulting formula to get the commutator [N(A),a^*(p)]. Then use this to compute the commutator formula [N(A),N(B)]=N([A,B]), using the rule [a,bc]=[a,b]c+b[a,c]. Finally specialize.